Is the space of smooth germs on a manifold finite dimensional or infinite-dimensional?

Question

Is $C^\infty_p$ a finite dimensional vector space or infinite dimensional, for any point $p$ in a manifold $M$?

The notation means the space of smooth germs.

Answer 1

For sake of easiness suppose that $M=\mathbb{R}$ and $p=0.$

Let $n$ be a natural number. We want to show that $n<\dim C^\infty_p.$

So consider functions $1,x,x^2,\dots,x^n$ and take their germs $[1],[x],[x^2],\dots,[x^n].$ We will prove that $[1],[x],[x^2],\dots,[x^n]$ are lineary independent.

Suppose that $a_0[1]+a_1[x]+a_2[x^2]+\dots+a_n[x^n]=[0].$ This means that $[a_0+a_1x+a_2x^2+\dots+a_nx^n]=[0].$ And by definition of germ, there exists some open set $U=(-\varepsilon,\varepsilon)$ such that $a_0+a_1x+a_2x^2+\dots+a_nx^n=0$ on $U.$ This means that $$\forall(x\in (-\varepsilon,\varepsilon))\hspace{10pt} a_0+a_1x+a_2x^2+\dots+a_nx^n=0.$$ Lets denote the left hand side by $f:(-\varepsilon,\varepsilon)\rightarrow\mathbb{R}.$ We see that $f(0)=0.$ Hence $a_0=0.$ We see that $f'(0)=0.$ Hence $a_1=0.$ Generally $f^{(k)}(0)=0=a_k$ holds for $k\leqslant n.$ And it means exactly that $[1],[x],[x^2],\dots,[x^n]$ are lineary independent. Hence $n<\dim C^\infty_p.$

Since $n$ was arbitrary, we get that $\dim C^\infty_p=\infty.$

On arbitrary manifold $M$ consider just functions, which around point $p$ takes form $1, \phi_1,\phi_1^2,\dots,\phi_1^n$ where $\phi:M\supset V\rightarrow\phi(V)\subset\mathbb{R}^n=(\phi_1,\dots,\phi_n)$ is some map around $p$ such that $\phi(p)=\theta.$